Empirical Formulas Molecular Formulas Since the formula of

Empirical Formulas Molecular Formulas

Since the formula of a compound indicates the number of atoms in the compound, CH 4 = 4 H atoms and 1 C atom we must first convert the masses of the elements to number of atoms g moles atoms (ratio)

Step 1: g moles Given: 38. 67% Carbon by mass 38. 67 g Carbon per 100 g of the compound Given: 16. 22% Hydrogen by mass 16. 22 g Hydrogen per 100 g of the compound Given: 45. 11% Nitrogen by mass

Step 1: g moles Given: 38. 67% Carbon by mass 38. 67 g Carbon per 100 g of the compound 38. 67 g C 1 mol C = 3. 220 mol C 12. 01 g C

Step 1: g moles • Given: 16. 22% Hydrogen by mass 16. 22 g Hydrogen per 100 g of the compound 16. 22 g H 1 mol H = 16. 09 mol H 1. 008 H

Step 1: g moles Given: 45. 11% Nitrogen by mass 45. 11 g Nitrogen per 100 g of the compound 45. 11 g N 1 mol N = 3. 219 mol N 14. 01 g N

Step 2: Which number is the smallest? 38. 67 g C 1 mol C = 3. 220 mol C 12. 01 g C 16. 22 g H 1 mol H = 16. 09 mol H 1. 008 H 45. 11 g N 1 mol N = 3. 219 mol N 14. 01 g N

Step 3: Whole Number Ratio • The smallest number from step two becomes the denominator when solving for atoms • C: 3. 220 = 1. 000 = 1 3. 220 • H: 16. 09 = 4. 997 = 5 3. 220 • N: 3. 219 = 1. 000 = 1 3. 220 CH 5 N

Molecular Formula Molar Mass (given) = Molecular Formula Multiplier Empirical Formula Mass Given: Cl. CH 2 has a molar mass of 98. 96 g/mol and an empirical formula mass of 49. 48 g/mol MM = 98. 69 g/mol EFM =2 49. 48 g/mol We now multiple (Cl. CH 2) by 2 =

Step 3: Whole Number Ratio • Example from Empirical Formula Notes Can find molar mass CH 5 N C – 1 x 12. 01 H – 5 x 1. 008 N – 1 x 14. 01 Compare to given (31. 06 g/mol) and divide 31. 06 (found molar mass) = 1, 1 = multiplier